Question

consider a pirof straight lies 3x2+xy+ky^2-5x+5y+2=0 and a line L: x+my+2=0, if there are exactly 2 distinct circles which touch all 3 lines then m=a. if there are 4 such circles then mnot equal to b. if there are no circles with non zero radius that touch all3 lines m=c. find sum of all possible values of k,a,b,c

Answer 1

2

Answer 2

The given quadratic represents two real lines if the discriminant of the quadratic in the “direction–variable” $m$ is $1-12k>0$ so that one must have $k<\tfrac1{12}$; in our answer $k=-2$ qualifies.

Centers of circles tangent to $L_1$ and $L_2$ lie on their two angle–bisectors.

Writing the tangency condition with the third line $L$ (with equation $x+my+2=0$) produces a quadratic in the distance along a chosen bisector; the “generic” four solutions reduce to two when the coefficient of the quadratic vanishes (this forces $m=a$ or, equivalently, if $m$ takes the other special value then one gets only 4 solutions provided $m\neq b$).

Finally, if the line $L$ happens to pass through the vertex (found by “completing the square” or by “differentiation” and elimination) then one proves that no circle with non–zero radius is possible; this is equivalent to $m=c$.

The unique solution is:

$$k=-2,\quad a=2,\quad b=-1,\quad c=3.$$

Hence, the sum is

$$k+a+b+c=(-2)+2+(-1)+3=2.$$